Question

How do I estimate the number of tennis balls that would fit in a classroom? I have to make assumptions and plan (sketch, equations to be used, etc). PLEASE HELP!

Answer 1

find the volume of the room find the volume of a tennis ball

Answer 2

what if I don't know the volume of the room or the volume of a tennis ball?

Answer 3

create variables?

Answer 4

I'm supposed to estimate and make assumptions...

Answer 5

Assume a tennis ball's diameter is 2.5 inches (it's really 2.62 but whatever which means it will take up 2.5 square inches.

Assume a room size of 20ft x 20ft which is 240 inches by 240 inches.

240^2 gives you the size of the room in square inches...just divided that by 2.5^2 to get 9216 tennis balls.

Answer 6

assuming your room is a box and now a circle, V=WLH volume of a tennisball you could find out

Answer 7

The volume of a tennis ball is really irrelevant since it will take up a square (2.62 in^2) when placed in a room

Answer 8

but to be precise

Answer 9

btw are you using area as that is only 2 dimensional... assuming the room's ceiling is on the floor

Answer 10

what do you mean by 2-dimensional?

Answer 11

Well...that's assuming you're going to fill up empty space in between each ball which you can't do unless you compress them...which would be whole other problem involved compressibility!

Answer 12

Crap..it should have been 240^3..you got me on that :)

Answer 13

exactly! but if she wants to fill a room full of tennisballs you'd want to use volume. if she wanted to fill the floor with tennisballs it'd be area.... i'm sure this is not a compressibility problem

Answer 14

and 2.5^3

Answer 15

do half tennisballs count?

Answer 16

and volume of a tennisball would be right also ... so volume = 4/3 PI * R^3

Answer 17

no half tennisballs don't count but I'm supposed to fill the whole room, therefore I should use volume? How do I create the equations/formulas?

Answer 18

Ok...I corrected the square vs cubic inch mistake :)

Assume a tennis ball's diameter is 2.5 inches (it's really 2.62 but whatever). Therefore it will take up 2.5 CUBIC inches (corrected).

Assume a room size of 20ft x 20ft which is 240 inches by 240 inches by 240 inches. 240^3 gives you the volume of the room in cubic inches...just divided that by 2.5^3 to get 884736 tennis balls.

\[240^3/2.5^3=884736\]

Answer 19

I still say the spherical volume doesn't matter...it's going to fill up a square in the room unless you compress them

Answer 20

they aren't stacking on top of eachother?

Answer 21

think of tennisballs much like a liquid in chemistry. they aren't just going to stack on top of eachother. they will fill spaces in between

Answer 22

the approximation of your answer will be close to mine though probably

Answer 23

Uncompressed, they will take up 2.5in^3...whether they are stacked or not...I don't see where the confusion is.

Answer 24

In real life, there would be some compression but that would of course be a deeper problem than what he's being asked IMO

Answer 25

where is your 2.5 cubic inches coming from... and sorry millie about this

Answer 26

The diameter (effective length) of a tennis ball cubed.

Answer 27

and what is that suppose to stand for.... area, volume, i'm not quite understanding why you are cubing it what equation are you using

Answer 28

ohhh, i see.. I'm only asked to estimate so yeah.

Answer 29

volume of a cube :P

Answer 30

true but is a ball a cube... my approximation will be within a higher degree of accuracy

Answer 31

Well, I politely disagree :) I guess he should add another assumption that the balls won't compress as they get stacked...which would be more likely if he just shrunk the classroom to 10x10x10 :P

Answer 32

My estimates are essentially based on the bounding (square) volume of the ball...not the spherical volume...since there will be space between the balls...that's all I'm saying

Answer 33

He can feel free to use spherical volume if he wishes...but I still think my result would be closer

Answer 34

o.o

Answer 35

either way millie you can do it his way or mine

V=LWH for your room
Volume of your sphere will equal 4/3*PI R^3 soo your equation will look something like this if done my way


\[\frac{Volume of Room}{\frac{4\pi}{3(2.63^3)}}\]

Answer 36

or \[\frac {Volume of room}{\frac{4}{3}\pi*(2.63)^3}\]

Answer 37

this will give you an approximation of how many tennisballs in any box/rectangular room

Answer 38

we could also add compression if you so desire storm lol

Answer 39

o yeah!! and round the number down as you can't have half a tennisball

Answer 40

I stand by my balls

Answer 41

Ok, I can't let this go...it's just not in my nature :)

Let's say you have a box that's 1000 cubic inches (10x10x10) and marbles which are all 1 cubic inch. Clearly, you can only fit 1000 marbles in the box. However, by your formula, you'd be able to fit ~1910 marbles.

The difference is large and is obviously due to the wasted space between the marbles. So when I was talking about tennis balls, I assume they are not being compressed and will fit the same formula. If they were compressed the problem becomes significantly more complicated as they will compress differently depending on how many balls are above each one.

One last note, if you use your formula, you have to remember that 2.63 is the diameter of the ball so you'd only half of that in your formula.

Answer 42

and actually it wouldn't be that it would be

\[\frac{volume of room}{\frac{4}{3}\pi(1.315)^3}\]

Answer 43

umm clearly if you get 1 cubic inch for the marble that is the volum eof the sphere and not a cube of the sphere

Answer 44

I noticed it after i came back... we kept using 2.63 so

Answer 45

A cubic inch is a cubic inch....the ball doesn't have to fill the cube but each side of the cube would be 1 inch in length.

Answer 46

I just don't think I'm explaining this well enough...it's clear as day to me

Answer 47

Maybe from a different angle: the room volume is 100% of the volume in the room. The ball's spherical volume doesn't account for the wasted space around the ball which takes away from how many balls you can fit into the room when you divide the room volume by the spherical volume.

My cubic inch marbles take up exactly 1 cubic inch because I'm using an imaginary bounding box around each of ball of 1 inch in diameter. If however we were use the volume of the sphere formula on a 1inch marble, you'd get ~.5 cubic inches making it appear to take up less space than it actually does. All of which makes perfect since which you consider that your formula would have fit ~1910 marbles in the same space I say would only be able to hold 1000 marbles.

Answer 48

your method is like taking blocks and building up. Basically like a toddler trying to make a square am i right? However that Volume of the cube could be a hugeeeeee depending on the significance for the room. so say you take a diameter of 2 like earlier a volume of a cube with side 2 will equal 8 as a volume of a ball will equal 4.1887.... significant difference if you are trying to fit a ton of balls into one spot.. take 4 for diameter your square will pull off a womping volume of 64 while the sphere is a simple 33.5

Answer 49

your answer will be close but not as accurate as perhaps finding the actual volume of the room to the actual volume of the ball and then dividing it and rounding down to get rid of all half balls

Answer 50

Yes, it's like a toddler stacking squares...exactly. How does your method account for the wasted space around the balls? Each ball effectively takes up a square since other balls cannot encroach in each other's bounding cube

Answer 51

it doesn't however yours just simply fills it with imaginary matter

Answer 52

it's not imaginary matter...it's space that cannot be filled by tennis balls

Answer 53

but the spacing between each ball is not exactly the same

Answer 54

Why wouldn't the be the same? It may not be **exactly** but it should be close enough for this exercise. Again, my assumption here is that all of the tennis balls are exactly the same size and aren't being compressed when they are packed into the room.

My point is simply that a 1inch box and a 1 inch in diameter ball will consume the same amount of *usable* space in this particular problem...

Answer 55

I guess in the end your method will be very accurate if you smush the balls into squares so they can be packed in tightly